Maximize Bitwise AND of first element with complement of remaining elements for any permutation of given Array

Given an array arr[] consisting of N integers, the task is to find the maximum value of Bitwise AND of the first element with the complement of remaining elements for any permutation of this array, i.e. 

A1 &(~A2) & (~A3) & ……& (~An)

Examples:

Input: arr[] = {1, 2, 4, 8, 16} 
Output: 16 
Explanation: 
For the permutation {16, 1, 2, 4, 8}, the maximum value of the expression can be obtained.

 



Input: arr[] = {0, 2, 3, 4, 9, 8} 
Output:
Explanation: 
For the permutation {4, 8, 9, 3, 2, 0}, the maximum value of the expression can be obtained

Naive Approach: The simplest approach to solve the problem is to generate all possible permutations of the given array and find the required value for each permutation and print the maximum among them. 
Time Complexity: O(N * N!) 
Auxiliary Space: O(N)
 

Efficient Approach: The above approach can be optimized by the following observation:  

  • The expression A1 &(~A2) & (~A3) & …… & (~An) is solely dependent on the value of A1. 
  • Therefore, to get the maximize the value from the expression, choose A1 such that it has the set bit of as highest significance as possible, which is unset in all other array elements.
  • Since, the order of remaining array elements doesn’t matter, print any permutation having the obtained A1 as the first element. 

Illustration: 
For arr[] = {1, 2, 4, 8, 16}
Binary representation of the array elements: 
(16)10 = (10000)2 
(8)10 = (01000)2 
(4)10 = (00100)2 
(2)10 = (00010)2 
(1)10 = (00001)2
As it can be seen that 16 has the highest significant set bit which is unset in all other array elements. Therefore, the required permutation of given permutation will contain 16 as the first element.
Hence, the required Bitwise AND is maximum for permutation having 16 as the first element, which is equal to 16 in this case. 

Below is the implementation of the above approach: 

C++

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// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
#define size_int 32
 
// Function to maximize the value for
// the given function and the array elements
int functionMax(int arr[], int n)
{
    // Vector array to maintain which bit is set
    // for which integer in the given array by
    // saving index of that integer
    vector<int> setBit[32];
 
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < size_int; j++) {
 
            // Check if j-th bit is set for
            // i-th integer
            if (arr[i] & (1 << j))
 
                // Push the index of that
                // integer in setBit[j]
                setBit[j].push_back(i);
        }
    }
 
    // Find the element having
    // highest significant set bit
    // unset in other elements
    for (int i = size_int; i >= 0; i--) {
        if (setBit[i].size() == 1) {
 
            // Place that integer at 0-th index
            swap(arr[0], arr[setBit[i][0]]);
            break;
        }
    }
 
    // Store the maximum AND value
    int maxAnd = arr[0];
    for (int i = 1; i < n; i++) {
        maxAnd = maxAnd & (~arr[i]);
    }
 
    // Return the answer
    return maxAnd;
}
 
// Driver Code
int main()
{
 
    int arr[] = { 1, 2, 4, 8, 16 };
    int n = sizeof arr / sizeof arr[0];
 
    // Function call
    cout << functionMax(arr, n);
 
    return 0;
}

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Java

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// Java Program to implement
// the above approach
import java.util.*;
class GFG{
 
static final int size_int = 32;
 
// Function to maximize the value for
// the given function and the array elements
static int functionMax(int arr[], int n)
{
    // Vector array to maintain which bit is set
    // for which integer in the given array by
    // saving index of that integer
    Vector<Integer> []setBit = new Vector[32 + 1];
    for (int i = 0; i < setBit.length; i++)
        setBit[i] = new Vector<Integer>();
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < size_int; j++)
        {
 
            // Check if j-th bit is set for
            // i-th integer
            if ((arr[i] & (1 << j)) > 0)
 
                // Push the index of that
                // integer in setBit[j]
                setBit[j].add(i);
        }
    }
 
    // Find the element having
    // highest significant set bit
    // unset in other elements
    for (int i = size_int; i >= 0; i--)
    {
        if (setBit[i].size() == 1)
        {
 
            // Place that integer at 0-th index
            swap(arr, 0, setBit[i].get(0));
            break;
        }
    }
 
    // Store the maximum AND value
    int maxAnd = arr[0];
    for (int i = 1; i < n; i++)
    {
        maxAnd = maxAnd & (~arr[i]);
    }
 
    // Return the answer
    return maxAnd;
}
   
static int[] swap(int []arr, int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}
   
// Driver Code
public static void main(String[] args)
{
 
    int arr[] = { 1, 2, 4, 8, 16 };
    int n = arr.length;
 
    // Function call
    System.out.print(functionMax(arr, n));
}
}
 
// This code is contributed by PrinciRaj1992

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Python3

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# Python 3 Program to
# implement the above approach
 
# Function to maximize the
# value for the given function
# and the array elements
def functionMax(arr, n):
   
    # Vector array to maintain
    # which bit is set for which
    # integer in the given array by
    # saving index of that integer
    setBit = [[] for i in range(32)]
 
    for i in range(n):
        for j in range(32):
           
            # Check if j-th bit is
            # set for i-th integer
            if (arr[i] & (1 << j)):
               
                # Push the index of that
                # integer in setBit[j]
                setBit[j].append(i)
 
    # Find the element having
    # highest significant set bit
    # unset in other elements
    i = 31
     
    while(i >= 0):
        if (len(setBit[i]) == 1):
           
            # Place that integer
            # at 0-th index
            temp = arr[0]
            arr[0] = arr[setBit[i][0]]
            arr[setBit[i][0]] = temp
            break
        i -= 1
 
    # Store the maximum
    # AND value
    maxAnd = arr[0]
    for i in range(1, n, 1):
        maxAnd = (maxAnd & (~arr[i]))
 
    # Return the answer
    return maxAnd
 
# Driver Code
if __name__ == '__main__':
    arr = [1, 2, 4, 8, 16]
    n = len(arr)
 
    # Function call
    print(functionMax(arr, n))
 
# This code is contributed by bgangwar59

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C#

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// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
static readonly int size_int = 32;
 
// Function to maximize the value for
// the given function and the array elements
static int functionMax(int []arr, int n)
{
    // List array to maintain which bit is set
    // for which integer in the given array by
    // saving index of that integer
    List<int> []setBit = new List<int>[32 + 1];
    for (int i = 0; i < setBit.Length; i++)
        setBit[i] = new List<int>();
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < size_int; j++)
        {
 
            // Check if j-th bit is set for
            // i-th integer
            if ((arr[i] & (1 << j)) > 0)
 
                // Push the index of that
                // integer in setBit[j]
                setBit[j].Add(i);
        }
    }
 
    // Find the element having
    // highest significant set bit
    // unset in other elements
    for (int i = size_int; i >= 0; i--)
    {
        if (setBit[i].Count == 1)
        {
 
            // Place that integer at 0-th index
            swap(arr, 0, setBit[i][0]);
            break;
        }
    }
 
    // Store the maximum AND value
    int maxAnd = arr[0];
    for (int i = 1; i < n; i++)
    {
        maxAnd = maxAnd & (~arr[i]);
    }
 
    // Return the answer
    return maxAnd;
}
   
static int[] swap(int []arr, int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}
   
// Driver Code
public static void Main(String[] args)
{
 
    int []arr = { 1, 2, 4, 8, 16 };
    int n = arr.Length;
 
    // Function call
    Console.Write(functionMax(arr, n));
}
}
 
// This code is contributed by Rajput-Ji

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Output: 

16







 

Time Complexity: O(N * sizeof(int)), where sizeof(int) is 32
Auxiliary Space: O(N)

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